Question

A solid sphere of radius $R$ starts rotating on rough horizontal surface with translational velocity $V_0$ and initial angular velocity ${\omega }_0=\frac{2V_0}{3R}$ . The sphere starts pure rolling after some time $t$. if $V$ is the translational velocity at pure rolling. Assume uniformly accelerated motion up to start of pure rolling:
if $S$ is the distance covered by the sphere in rotational motion up to the instant at which pure rolling starts, find the velocity (translational) of the sphere during the pure rolling

• A. $\frac{33}{38}\frac{S}{t}$
• B. $\frac{31}{38}\frac{S}{t}$
• C. $\frac{38}{33}\frac{S}{t}$
• D. $\frac{37}{38}\frac{S}{t}$

Answer 1

The correct option is C $\frac{38}{33}\frac{S}{t}$

Draw a diagram of given problem.

Find angular acceleration of the sphere.
Formula used:$\omega ={\omega }_0+\alpha t$
Given,
${\omega }_0=\frac{2V_0}{3R}$

$V_0=\frac{3}{2}R{\omega }_0$
according to diagram, angular momentum is conserved about $A,$
$L_1=L_2$
Therefore,
$\frac{2}{5}m{R}^2{\omega }_0+m{V}_{0}R=\frac{2}{5}m{R}^2\omega +mVR$

$\frac{2}{5}m{R}^2\times \frac{2V_0}{3R}+m{V}_{0}R=\frac{2}{5}m{R}^2\frac{V}{R}+mVR$

$V_0=\frac{21}{19}V\left(i\right)$
Now, according to question $\omega$ and $V$ are translational velocity for pure rolling starts,
$\omega ={\omega }_0+\alpha t$
From equation $\left(i\right)$,we get
$\alpha =\frac{5V}{19Rt}$
Where, $t$ is time when pure rolling starts.

Find angle by which sphere rotates.
Formula used:$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$
Pure rolling starts when sphere start rotated by an angle $\theta$,
Therefore,
$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

$\theta =\frac{2V_0}{3R}t+\frac{1}{2}.\frac{5V}{19Rt}.t^2$

$\theta =\frac{33}{38}\frac{Vt}{R}$

$R\theta =S=\frac{33}{38}Vt$
Therefore,
Translational velocity$\left(V\right),$
$V=\frac{38}{33}\frac{S}{t}$

Final answer: (d)